undifined 1/0 ?!!!

**avemt1** · Nov14-03, 02:48 PM

why is anything over zero undefined?
This is a question I have faced for a while and have not found an answer.
Could you please help me?

**NateTG** · Nov14-03, 04:25 PM

It depends on how you define division.

Usually
a/b=x
means that
x*b=a

Now, let's take a look at the case where b=0 and a=0;
x*0=0.

Clearly any x works.

And in the case where b=0 and a is not zero:
x*0=a
Clearly no x works.

Either way, there is no unique x so that the equation works, so it remains undefined.

**HallsofIvy** · Nov15-03, 09:10 AM

By the way, because of the distinction that NateTG noted,
any x satisfies x*0= 0 but no x satisfies x*0= b for b non-zero,

it is common to say that 0/0 is "undetermined" while b/0, for b non-zero, is "undefined".

**Hessam** · Dec21-03, 01:14 PM

because... 1/ .01 = 100 and 1/ .00000001 = 1000000000 and so forth... and thus as the numbers get smaller on the bottom, the answer becomes higher... so much that 1/0 = undefined

**HallsofIvy** · Dec21-03, 01:39 PM

posted by Hessam
I thought it was...because... 1/ .01 = 100 and 1/ .00000001 = 1000000000 and so forth... and thus as the numbers get smaller on the bottom, the answer becomes higher... so much that 1/0 = undefined

You were mistaken. If I define f(x) to be 1/x if x is not 0 and 1 if x=0, then it is also true that f(.1), f(.01), f(.00000001), etc get larger and larger but f(0) is not "undefined" (f is merely "discontinuous" at 0). 1/0 is "undefined" because there is no way to define it that does not violate some basic property of the real numbers and the definition of "/".

**suyver** · Dec22-03, 04:00 AM

Originally posted by Hessam
because... 1/ .01 = 100 and 1/ .00000001 = 1000000000 and so forth... and thus as the numbers get smaller on the bottom, the answer becomes higher... so much that 1/0 = undefined

At the same time, I can say that:
because... 1/ -.01 = -100 and 1/ -.00000001 = -1000000000 and so forth... and thus as the numbers get smaller on the bottom, the answer becomes lower... so much that 1/0 = undefined

More formally,
$LaTeX Code: \\lim_{x\\uparrow 0}\\frac{1}{x}\\rightarrow -\\infty$
but also
$LaTeX Code: \\lim_{x\\downarrow 0}\\frac{1}{x}\\rightarrow +\\infty$

That's bad news! For me, that's the simplest argument for saying that 1/0 is undefined. Note that I can also get other answers, if you'd like I can probably find a limit such that
$LaTeX Code: \\lim_{x\\rightarrow 0}\\frac{1}{x}\\rightarrow -\\sqrt{\\frac{i\\pi}{e}}$ [:D]

**HallsofIvy** · Dec22-03, 07:19 AM

Why in the world would that be simpler? The original question is about basic arithmetic and has nothing to do with limits.

1/0 is undefined because 1/0= x is equivalent to 1= 0*x which is not true for any x.

I don't see how using limits on some specific sequence would be simpler than that!

**suyver** · Dec22-03, 07:30 AM

I guess it's a question of mindset and what you're used to. I find my argument simpler / clearer than yours, HallsofIvy, though I directly acknowledge that (many) others might disagree. That is why I included this argument: if people might have problems with your or NateTG's argument, then maybe the one I provided will give some them insight. If not, then no harm done. However, if you think it's stupid, then please feel free to remove my posts.

**HallsofIvy** · Dec22-03, 02:18 PM

Well, if you find limits easier than multiplication, you have a remarkable mind!

**NateTG** · Dec22-03, 06:08 PM

Originally posted by HallsofIvy
Well, if you find limits easier than multiplication, you have a remarkable mind!

In some sense, limits are a much simpler concept, even if the formalisms associated with it are a bit more involved.

On a conceptual level, I would say that limits are much simpler than multiplication.

In practice, multiplication is much more usefull.

**avemt1** · Jan5-04, 11:57 AM

because... 1/ .01 = 100 and 1/ .00000001 = 1000000000 and so forth... and thus as the numbers get smaller on the bottom, the answer becomes higher... so much that 1/0 = undefined

and 1/infinity is equal to 1/0
1/-infinity is equal to 1/0
if i am incorrect please correct me

**HallsofIvy** · Jan5-04, 01:17 PM

No, I wouldn't say you are incorrect but you are not very precise (and in mathematics precision is essential!).

"infinity" is not a standard real number and if you are talking about the real numbers, "1/infinity= 0" and "1/(-infinity)= 0" are short hand for "the limit of 1/x as x goes to infinity is 0" and "the limit of 1/x as x goes to -infinity is 0".

If you are using one of the several "extended number systems" in which infinity is defined, then you should say so.

That is one reason why it is not very good mathematics to say "1/0= infinity".

**avemt1** · Jan5-04, 02:56 PM

"infinity" is not a standard real number and if you are talking about the real numbers, "1/infinity= 0" and "1/(-infinity)= 0" are short hand for "the limit of 1/x as x goes to infinity is 0" and "the limit of 1/x as x goes to -infinity is 0".

So if zero is the equivalent of infinity and -infinity then the numberline as a whole is not just a straight line, but a loop that resembles the sign for infinity. (KUNUNDRUM?)

If you are using one of the several "extended number systems" in which infinity is defined, then you should say so.

I am treating infinity as a variable, and the number system explained and described above.

**NateTG** · Jan5-04, 03:06 PM

It's quite possible to do math with $LaTeX Code: \\infty$ but it lacks some of the properties that are normally associated with numbers.

For example, $LaTeX Code: \\intfy-\\infty$ and $LaTeX Code: \\frac{\\infty}{\\infty}$ will cause problems.

**avemt1** · Jan7-04, 09:35 AM

It's quite possible to do math with but it lacks some of the properties that are normally associated with numbers.

I see no problem with using negative infinity, but i do see what you mean for the tricky infinity, because if you multiply infinity by its tricky one then it will turn out as the tricky one.
infinity multiplied by itself commes out as infinity. This is the problem with working within the boudaries of infinities, you have no room to move. That is why infinities do not work to explain the world around us.
I do understand this consept, as you can see, but i do not understand using negative infinity as a problem.

**HallsofIvy** · Jan7-04, 09:52 PM

Would you mind telling us what in the world you mean by the "tricky infinity"?

By the way, you copied what I said about infinity not being in the standard real numbers and then said

So if zero is the equivalent of infinity and -infinity then the numberline as a whole is not just a straight line, but a loop that resembles the sign for infinity.

How in the world did you get that "zero is the equivalent of infinity and - infinity" from what I said? No, zero is not the equivalent of infinity in any sense!